3.2.37 \(\int \frac {A+B x}{x^3 (b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=166 \[ \frac {256 c^3 (b+2 c x) (3 b B-4 A c)}{63 b^7 \sqrt {b x+c x^2}}-\frac {32 c^2 (b+2 c x) (3 b B-4 A c)}{63 b^5 \left (b x+c x^2\right )^{3/2}}+\frac {4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac {2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {792, 658, 614, 613} \begin {gather*} \frac {256 c^3 (b+2 c x) (3 b B-4 A c)}{63 b^7 \sqrt {b x+c x^2}}-\frac {32 c^2 (b+2 c x) (3 b B-4 A c)}{63 b^5 \left (b x+c x^2\right )^{3/2}}+\frac {4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac {2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^3*(b*x + c*x^2)^(5/2)),x]

[Out]

(-2*A)/(9*b*x^3*(b*x + c*x^2)^(3/2)) - (2*(3*b*B - 4*A*c))/(21*b^2*x^2*(b*x + c*x^2)^(3/2)) + (4*c*(3*b*B - 4*
A*c))/(21*b^3*x*(b*x + c*x^2)^(3/2)) - (32*c^2*(3*b*B - 4*A*c)*(b + 2*c*x))/(63*b^5*(b*x + c*x^2)^(3/2)) + (25
6*c^3*(3*b*B - 4*A*c)*(b + 2*c*x))/(63*b^7*Sqrt[b*x + c*x^2])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}--\frac {\left (2 \left (-3 (-b B+A c)-\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {1}{x^2 \left (b x+c x^2\right )^{5/2}} \, dx}{9 b}\\ &=-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}-\frac {(10 c (3 b B-4 A c)) \int \frac {1}{x \left (b x+c x^2\right )^{5/2}} \, dx}{21 b^2}\\ &=-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}+\frac {\left (16 c^2 (3 b B-4 A c)\right ) \int \frac {1}{\left (b x+c x^2\right )^{5/2}} \, dx}{21 b^3}\\ &=-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac {32 c^2 (3 b B-4 A c) (b+2 c x)}{63 b^5 \left (b x+c x^2\right )^{3/2}}-\frac {\left (128 c^3 (3 b B-4 A c)\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{63 b^5}\\ &=-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac {32 c^2 (3 b B-4 A c) (b+2 c x)}{63 b^5 \left (b x+c x^2\right )^{3/2}}+\frac {256 c^3 (3 b B-4 A c) (b+2 c x)}{63 b^7 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 145, normalized size = 0.87 \begin {gather*} \frac {6 b B x \left (-3 b^5+6 b^4 c x-16 b^3 c^2 x^2+96 b^2 c^3 x^3+384 b c^4 x^4+256 c^5 x^5\right )-2 A \left (7 b^6-12 b^5 c x+24 b^4 c^2 x^2-64 b^3 c^3 x^3+384 b^2 c^4 x^4+1536 b c^5 x^5+1024 c^6 x^6\right )}{63 b^7 x^3 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^3*(b*x + c*x^2)^(5/2)),x]

[Out]

(6*b*B*x*(-3*b^5 + 6*b^4*c*x - 16*b^3*c^2*x^2 + 96*b^2*c^3*x^3 + 384*b*c^4*x^4 + 256*c^5*x^5) - 2*A*(7*b^6 - 1
2*b^5*c*x + 24*b^4*c^2*x^2 - 64*b^3*c^3*x^3 + 384*b^2*c^4*x^4 + 1536*b*c^5*x^5 + 1024*c^6*x^6))/(63*b^7*x^3*(x
*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.48, size = 163, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-7 A b^6+12 A b^5 c x-24 A b^4 c^2 x^2+64 A b^3 c^3 x^3-384 A b^2 c^4 x^4-1536 A b c^5 x^5-1024 A c^6 x^6-9 b^6 B x+18 b^5 B c x^2-48 b^4 B c^2 x^3+288 b^3 B c^3 x^4+1152 b^2 B c^4 x^5+768 b B c^5 x^6\right )}{63 b^7 x^5 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^3*(b*x + c*x^2)^(5/2)),x]

[Out]

(2*Sqrt[b*x + c*x^2]*(-7*A*b^6 - 9*b^6*B*x + 12*A*b^5*c*x + 18*b^5*B*c*x^2 - 24*A*b^4*c^2*x^2 - 48*b^4*B*c^2*x
^3 + 64*A*b^3*c^3*x^3 + 288*b^3*B*c^3*x^4 - 384*A*b^2*c^4*x^4 + 1152*b^2*B*c^4*x^5 - 1536*A*b*c^5*x^5 + 768*b*
B*c^5*x^6 - 1024*A*c^6*x^6))/(63*b^7*x^5*(b + c*x)^2)

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fricas [A]  time = 0.41, size = 177, normalized size = 1.07 \begin {gather*} -\frac {2 \, {\left (7 \, A b^{6} - 256 \, {\left (3 \, B b c^{5} - 4 \, A c^{6}\right )} x^{6} - 384 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{5} - 96 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} + 16 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{3} - 6 \, {\left (3 \, B b^{5} c - 4 \, A b^{4} c^{2}\right )} x^{2} + 3 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} x\right )} \sqrt {c x^{2} + b x}}{63 \, {\left (b^{7} c^{2} x^{7} + 2 \, b^{8} c x^{6} + b^{9} x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

-2/63*(7*A*b^6 - 256*(3*B*b*c^5 - 4*A*c^6)*x^6 - 384*(3*B*b^2*c^4 - 4*A*b*c^5)*x^5 - 96*(3*B*b^3*c^3 - 4*A*b^2
*c^4)*x^4 + 16*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x^3 - 6*(3*B*b^5*c - 4*A*b^4*c^2)*x^2 + 3*(3*B*b^6 - 4*A*b^5*c)*x)*
sqrt(c*x^2 + b*x)/(b^7*c^2*x^7 + 2*b^8*c*x^6 + b^9*x^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(5/2)*x^3), x)

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maple [A]  time = 0.05, size = 158, normalized size = 0.95 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (1024 A \,c^{6} x^{6}-768 B b \,c^{5} x^{6}+1536 A b \,c^{5} x^{5}-1152 B \,b^{2} c^{4} x^{5}+384 A \,b^{2} c^{4} x^{4}-288 B \,b^{3} c^{3} x^{4}-64 A \,b^{3} c^{3} x^{3}+48 B \,b^{4} c^{2} x^{3}+24 A \,b^{4} c^{2} x^{2}-18 B \,b^{5} c \,x^{2}-12 A \,b^{5} c x +9 b^{6} B x +7 A \,b^{6}\right )}{63 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{7} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^3/(c*x^2+b*x)^(5/2),x)

[Out]

-2/63*(c*x+b)*(1024*A*c^6*x^6-768*B*b*c^5*x^6+1536*A*b*c^5*x^5-1152*B*b^2*c^4*x^5+384*A*b^2*c^4*x^4-288*B*b^3*
c^3*x^4-64*A*b^3*c^3*x^3+48*B*b^4*c^2*x^3+24*A*b^4*c^2*x^2-18*B*b^5*c*x^2-12*A*b^5*c*x+9*B*b^6*x+7*A*b^6)/x^2/
b^7/(c*x^2+b*x)^(5/2)

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maxima [A]  time = 0.51, size = 270, normalized size = 1.63 \begin {gather*} -\frac {64 \, B c^{3} x}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} + \frac {512 \, B c^{4} x}{21 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {256 \, A c^{4} x}{63 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{5}} - \frac {2048 \, A c^{5} x}{63 \, \sqrt {c x^{2} + b x} b^{7}} - \frac {32 \, B c^{2}}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} + \frac {256 \, B c^{3}}{21 \, \sqrt {c x^{2} + b x} b^{5}} + \frac {128 \, A c^{3}}{63 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} - \frac {1024 \, A c^{4}}{63 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {4 \, B c}{7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x} - \frac {16 \, A c^{2}}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3} x} - \frac {2 \, B}{7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x^{2}} + \frac {8 \, A c}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x^{2}} - \frac {2 \, A}{9 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

-64/21*B*c^3*x/((c*x^2 + b*x)^(3/2)*b^4) + 512/21*B*c^4*x/(sqrt(c*x^2 + b*x)*b^6) + 256/63*A*c^4*x/((c*x^2 + b
*x)^(3/2)*b^5) - 2048/63*A*c^5*x/(sqrt(c*x^2 + b*x)*b^7) - 32/21*B*c^2/((c*x^2 + b*x)^(3/2)*b^3) + 256/21*B*c^
3/(sqrt(c*x^2 + b*x)*b^5) + 128/63*A*c^3/((c*x^2 + b*x)^(3/2)*b^4) - 1024/63*A*c^4/(sqrt(c*x^2 + b*x)*b^6) + 4
/7*B*c/((c*x^2 + b*x)^(3/2)*b^2*x) - 16/21*A*c^2/((c*x^2 + b*x)^(3/2)*b^3*x) - 2/7*B/((c*x^2 + b*x)^(3/2)*b*x^
2) + 8/21*A*c/((c*x^2 + b*x)^(3/2)*b^2*x^2) - 2/9*A/((c*x^2 + b*x)^(3/2)*b*x^3)

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mupad [B]  time = 1.41, size = 266, normalized size = 1.60 \begin {gather*} \frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {4\,c^3\,\left (176\,A\,c-111\,B\,b\right )}{63\,b^5}+\frac {2\,c^3\,\left (247\,A\,c-138\,B\,b\right )}{63\,b^5}+\frac {b\,\left (\frac {184\,A\,c^5-96\,B\,b\,c^4}{63\,b^6}-\frac {4\,c^4\,\left (247\,A\,c-138\,B\,b\right )}{63\,b^6}\right )}{c}\right )+\frac {2\,c^2\,\left (176\,A\,c-111\,B\,b\right )}{63\,b^4}\right )}{x^2\,{\left (b+c\,x\right )}^2}-\frac {\sqrt {c\,x^2+b\,x}\,\left (18\,B\,b^3-52\,A\,b^2\,c\right )}{63\,b^6\,x^4}-\frac {\sqrt {c\,x^2+b\,x}\,\left (\frac {1024\,A\,c^4-768\,B\,b\,c^3}{63\,b^6}+\frac {2\,c\,x\,\left (1024\,A\,c^4-768\,B\,b\,c^3\right )}{63\,b^7}\right )}{x\,\left (b+c\,x\right )}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{9\,b^3\,x^5}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}\,\left (23\,A\,c-12\,B\,b\right )}{21\,b^5\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^3*(b*x + c*x^2)^(5/2)),x)

[Out]

((b*x + c*x^2)^(1/2)*(x*((4*c^3*(176*A*c - 111*B*b))/(63*b^5) + (2*c^3*(247*A*c - 138*B*b))/(63*b^5) + (b*((18
4*A*c^5 - 96*B*b*c^4)/(63*b^6) - (4*c^4*(247*A*c - 138*B*b))/(63*b^6)))/c) + (2*c^2*(176*A*c - 111*B*b))/(63*b
^4)))/(x^2*(b + c*x)^2) - ((b*x + c*x^2)^(1/2)*(18*B*b^3 - 52*A*b^2*c))/(63*b^6*x^4) - ((b*x + c*x^2)^(1/2)*((
1024*A*c^4 - 768*B*b*c^3)/(63*b^6) + (2*c*x*(1024*A*c^4 - 768*B*b*c^3))/(63*b^7)))/(x*(b + c*x)) - (2*A*(b*x +
c*x^2)^(1/2))/(9*b^3*x^5) - (2*c*(b*x + c*x^2)^(1/2)*(23*A*c - 12*B*b))/(21*b^5*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{3} \left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((A + B*x)/(x**3*(x*(b + c*x))**(5/2)), x)

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